# Engineering Physics Questions and Answers – Mass Variation with Velocity

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This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Mass Variation with Velocity”.

1. The lowest possible mass of a particle is its _____________
a) Relativistic mass
b) Inertial mass
c) Gravitational mass
d) Rest mass

Explanation: We know the relation, m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Thus, the minimum velocity one can acquire is zero. For that, the expression reduces to m = m0.
Thus, rest mass is the lowest possible mass of a particle.

2. The rest mass of a photon is equal to ___________
a) Gravitational mass
b) Relativistic mass
c) Inertial mass
d) Zero

Explanation: A photon is defined as a quantum of energy. It travels with the sped of light and is said to have no rest mass.

3. If the direction of velocity is changed by 180o simultaneously, the mass would _____________
a) Change simultaneously
b) Not Change
c) Increase
d) Decrease

Explanation: As we know, the relation for mass variation with velocity is m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$.
As we can see, it depends on the square of the velocity. Thus, changing the direction of velocity would not effect it.

4. Which basic law is used for the derivation of mass variation with velocity?
a) Law of conservation of Energy
b) Law of conservation of Kinetic Energy
c) Law of conservation of Momentum
d) Law of conservation of mass

Explanation: The law of conservation of momentum is used for the derivation of the formula. It is assumed that two particles in a frame collides elastically and are rebounded.

5. The speed of light is a limiting velocity, unattainable by a material body.
a) True
b) False

Explanation: The velocity of light is the limiting velocity. From the expression, m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$, as the velocity of the object approaches the speed of light, it’s relativistic mass approaches infinity. Thus, the speed of light is unattainable by a material body. Only those with zero rest mass, i.e., photons, can attain the speed of light.
Explanation: We know the relation m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Thus, according to it, the below graph shows the correct variation of mass with velocity. 7. At what speed the mass of an object would be double its value at rest?
a) √2 c
b) c/√2
c) √3c/2
d) 2c/√3

Explanation: Using the relation, m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Here, m = 2m0. Therefore, 2 m0. = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
4(1 – v2/c2) = 1
V = √3 c/2.

8. A man, who weighs 60 kg on earth, weighs 61 kg on a rocket, as measured by an observer on earth. What is the speed of the rocket?
a) 2.5 X 108 m/s
b) 2.5 X 107 m/s
c) 5.5 X 107 m/s
d) 5.5 X 108 m/s

Explanation: Now, as we know m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Here, m = 61 kg and m0 = 60 kg
Therefore, 61/60 = 1/√$$\sqrt{1-\frac{v^2}{c^2}}$$
v2/c2 = 1 – (60/61)2 = 121/3600
v = 11 X c/60 = 5.5 X 107 m/s.

9. An object of rest mass 6 Kg is moving with a speed of 0.8c. Its effective mass is ____________
a) 6 Kg
b) 8 Kg
c) 10 Kg
d) 12 Kg

Explanation: We know the relation, m = $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Now, here v = 0.8 c
Therefore, v2/c2 = 0.64
Hence, 1 -v2/c2 = 0.36
Thus, the expression reduces to = 6/0.6 Kg
= 10 Kg.

10. A body is initially at rest. It explodes into two objects of mass 1 Kg and 2 Kg moving with a speed of 0.6 times the speed of velocity each. What would be the mass of the original body at rest?
a) 2 Kg
b) 2.5 Kg
c) 3.5 Kg
d) 3.75 Kg

Explanation: E0 = mc2 = γm1c2 + γm2c2
E0 = $$\frac{m1C^2}{\sqrt{1-\frac{v^2}{c^2}}} + \frac{m2C^2}{\sqrt{1-\frac{v^2}{c^2}}}$$
m = E/c2
Therefore, m = $$\frac{m1+m2}{\sqrt{1-\frac{v^2}{c^2}}}$$
m = 3/0.8
m = 3.75 Kg.

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